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Math Help - A messy Indefinite Integral

  1. #1
    Lord of certain Rings
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    A messy Indefinite Integral

     \int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du

    While helping disclaimer with the differential equation problem, I ran into this integral. I am having a hard time with it. So please help.

    Thanks,
    Isomorphism
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Isomorphism View Post
     \int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du

    While helping disclaimer with the differential equation problem, I ran into this integral. I am having a hard time with it. So please help.

    Thanks,
    Isomorphism
    That's because it is a stinker to integrate. It looks like it's going to be done using a partial fraction expansion, but as the zeros of the denominator are complex I'm leaving it alone. (This is why I didn't respond to the other post.)

    -Dan
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    Lord of certain Rings
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    Quote Originally Posted by topsquark View Post
    That's because it is a stinker to integrate. It looks like it's going to be done using a partial fraction expansion, but as the zeros of the denominator are complex I'm leaving it alone. (This is why I didn't respond to the other post.)

    -Dan
    Ya I thought so. But does that mean there is a better substitution?

    P.S: The online integrator gave me a scary answer
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  4. #4
    Moo
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    Hello,

    I cheated a little and asked my TI...

    There is a real root and 2 complex roots...

    ... and they look awful
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    Lord of certain Rings
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    That looks much neater than the wolfram online integrators solution. However they have just neatly hidden the ugly ones in that big-Oh notation, isnt it?
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    Moo
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    Hello,

    Quote Originally Posted by Plato View Post
    \frac{{1 + u + u^2 }}<br />
{{1 - u - u^2  + u^3 }} = \frac{3}<br />
{{2\left( {u - 1} \right)^2 }} - \frac{3}<br />
{{4\left( {u - 1} \right)}} + \frac{1}<br />
{{4\left( {u + 1} \right)}}<br />
    But it's 1-u-u^2{\color{red}-}u^3
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  8. #8
    Lord of certain Rings
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    The integral in question is  \int \frac{1+u+u^2}{1 - u - u^2 - u^3} \, du

    Quote Originally Posted by Plato View Post
    \frac{{1 + u + u^2 }}<br />
{{1 - u - u^2  + u^3 }} = \frac{3}<br />
{{2\left( {u - 1} \right)^2 }} - \frac{3}<br />
{{4\left( {u - 1} \right)}} + \frac{1}<br />
{{4\left( {u + 1} \right)}}<br />
    So how will that help? Is there some manipulation I cannot see?
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  9. #9
    Moo
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    If it was +u^3 it would have been easier
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  10. #10
    Lord of certain Rings
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    Quote Originally Posted by Moo View Post
    If it was +u^3 it would have been easier
    Yes I know... I wished the same while working on it. I even rechecked the computations to see if I messed up with some sign.... But in vain.....
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  11. #11
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    I was going to comment on the fact that integrals shouldn't be classified as messy or nasty, but rather challenging.

    Then I saw what you were talking about, and I must concur with the adjective you used.

    EEEK!
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  12. #12
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    There's no way to compute this, you can't even factorise the denominator as a "nice" form.
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